Probability to Compute Divisor of a Hidden Integer

The article makes an investigation on the probability of finding the greatest common divisor between a given integer and a hidden integer that lies in an integer interval. It shows that, adding the integers that are picked randomly in the interval results in a much bigger probability than subtracting the picked integers one with another. Propositions and theorems are proved and formulas to calculate the probabilities are presented in detail. The research is helpful in developing probabilistic algorithm of integer factorization.


Introduction
Given an odd integer N that has the greatest common divisor (GCD) d with another odd integer e in a large odd interval that consists in consecutive odd numbers, what is the probability to find out d ?This question is close to the problem of integer factorization, as investigated in (Xingbo Wang, 2017(1)), (Jianhui Li, 2017), (Dongbo Fu, 2017) and (Xingbo Wang, 2017(2)).As stated in (Xingbo Wang, 2017(3)) the answer to the question relies on a detail study on the properties of odd integers on an odd interval.This article makes an investigation on the probability of finding the integer e in a large odd interval.The research shows that, by adding two or more terms contained in the interval, it has a very big probability to find out d.

Symbols and Notations
In this whole article, an odd interval [a, b] is a set of consecutive odd numbers that take a as lower bound and b as upper bound, for example, [3,11] = {3, 5, 7, 9, 11}.Symbol ⌊x⌋ is the floor function of real number x that satisfies ⌊x⌋ ≤ x < ⌊x⌋ + 1 .Symbol P(n, o) or P(S , o) is to express the total probability to find successfully an objective number o in a set S consisting in n terms and symbol P k (S , o) is the probability to find successfully o by picking randomly k terms in set S .Symbol is the binomial coefficient defined by

Lemmas
Lemma 1(See in Xingbo Wang, 2016& 2017(3)) Let p be a positive odd integer; then among p consecutive positive odd integers there exists one and only one that can be divisible by p.Let q be a positive odd number and S be a finite set that is composed of consecutive odd integers; then S needs at least (n − 1)q + 1 elements to have n multiples of q.If s α ∈ S is a multiple of o,then so it is with s α+o

Main Results and Proofs
Proposition 1 Let p be an odd number and n be a positive integer with n ≤ p; suppose s 1 , s 2 , • • • , s n are n consecutive odd numbers and S = {s j − s i |1 ≤ i < j ≤ n}; then there is not a p's multiple in S .
Proof.Referring to the proof of Theorem 2 in (Xingbo Wang, 2017(3)) immediately yields the conclusion.
Proposition 2 Suppose N is a composite odd integer and n is a positive integer; let S = {s 1 , s 2 , • • • , s n } be a set that consists in n consecutive odd integers that satisfy, for an m with Proposition 3 Let N be an odd number and n be a positive integer.
Proof.By Lemma 2 and Proposition 1, there are αn − α(α+1) 2 d multiples of d in the form of s j − s i ,hence it holds Theorem 1 Suppose N is a composite odd integer and n is a positive integer; let S = {s 1 , s 2 , ..., s n } be a set that consists in n consecutive odd integers that satisfy GCD(N, s i ) =  2.There are ways to obtain 4s m .In fact, choosing two consecutive terms, s m− j and s m− j−1 , and another two consecutive terms, s m+ j and s m+ j+1 , will result in 4s m with j = 1, 2, ..., m − 2. Likewise, arbitrary symmetricallydistributed four terms, say s m−k , s m−l , s m+k and s m+l , lead to 4s m .Hence there are ways to obtain 4s m .
4. There is one way to obtain s m , that is, to choose s m itself.
Consequently, the total ways of picking one to k left-terms, denoted by Λ k , that can produce multiples of s m are calculated by Now consider the case m ≤ n < 2m − 1.There are m − 1 left-terms but there are merely n − m right-terms, as shown in figure 1.
Proposition 5 Suppose N is a composite odd integer and n is a positive integer; let S = {s 1 , s 2 , • • • , s n } be a set that consists in n consecutive odd integers that satisfy GCD(N, s by picking terms in S and adding the picked terms together the probability P(S , d) is bigger than 1 n−1 .Proof.Referring to Theorem 1 and its proof, it holds Proposition 6 Suppose N is a composite odd integer and n is a positive integer; let S = {s 1 , s 2 , • • • , s n } be a set that consists in n consecutive odd integers that satisfy , for an m with 1 < m < n, GCD (N, s picking terms in S and adding the picked terms together the probability P(S , d) is Referring to the proofs of Theorem 1 and Theorem 2 and considering the case 1 < n − m ≤ m − 1 , one can see that, picking an arbitrary right-term or one more arbitrary right-terms among s m+1 , s m+2, ..., s n can always result in one or one more multiples of s m by adding the picked terms with their symmetric left-terms.Hence the whole events can be considered as follows.
1. Choose randomly 1 term in S , the probability P 1 (S , d) that a right-term s m+ j with j = 1, 2, ..., n − m is picked is calculated by 2. Choose randomly 2 terms in S , the probability P 2 (S , d) that two right-terms are picked is calculated by 3. Choose randomly k terms in S , the probability that k right-terms are picked is calculated by Choosing s m to obtain d leads to a probability P 1 (S , d) = 1 n .Consequently, the total probability P(S , d) is given by By Lemma 3, it yields When n = 2m − 1 , it yields x 3 (x+3) 3 . Since f ′ (3) = 0 and f ′′ (3) = − 1 27 , f (x) reaches its mmaximal value 1 3 when x = 3, and thus 1 − f (x) reaches its minimal value 2 3 .

Conclusions and Future Work
Finding the GCD between a given integer and a hidden integer in the sequence of integers plays an important role in solving the problems of factoring integers.There was successful approach like famous Pollards rho algorithm (Eric Bach, 1991) that applies integer minus another one to increase the probability of computing the GCD.Pollards rho method being applied on an infinite integer interval, its probability does increase to a relative big quantity.However, when the interval that contains the objective integer is limited to a finite range, as analyzed in (Xingbo Wang, 2017(1)), the subtraction of two terms in the interval cannot increase the probability as expected, as proved in Proposition 2. By inequality (1) in Proposition 2, when the GCD d is bigger enough, α = ⌊ n d

⌋
tends to be zero and the probability also tends to be zero.On the contrary, adding the terms in the interval, as Theorem 1 and Theorem 2 say, can increase a lot of probability.According to Proposition 6, if the objective number lies near the middle of the interval, the probability increase with n's increasing.These results provide a different way from the classical method of subtraction.This is the future work that is required to do: developing new probabilistic algorithm.Hope to see better achievements.
there are totally 2 n−m ways to compute d if m < n < 2m − 1, and there are totally 2 m−1 ways to compute d if n = 2m − 1 .Proof.When 1 < m < n, there always are terms on the left and the right of s m .For convenience, the terms on the left are called left-terms and the terms on the right are called right-terms.Since S consists in n consecutive odd integers, when n ≥ 2m − 1 it knows that, there are m − 1 left-terms and there are at least m − 1 right-terms.This time, it yields the following facts.1.There are ( m−1 1 ) = m − 1 ways to obtain 2s m by choosing s m− j and s m+ j that fit s m− j + s m+ j = 2s m with j = 1, 2, ..., m − 1.

Figure 1 .
Figure 1.Distribution of terms around s m